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Consider $n$ independent normally distributed random variables

$$X_i\sim N(\mu_i,\sigma_i^2)$$

and denote $Y = \max\limits_{1\leq i\leq n}\{X_i\}$. We can define the probabilities, for each $1\leq i\leq n$,

$$\tilde{P}_i = \mathbb{P}(X_i=Y)$$

where $\tilde{P}_i$ is a function of $\vec{\mu} = (\mu_1,\ldots,\mu_n)$ and $\vec{\sigma}=(\sigma_1,\ldots,\sigma_n).$ I'm interested in the inverse problem: given some observed probabilities $P_1,\ldots,P_n$ (summing to unity),

$$\min\limits_{(\vec{\mu},\vec{\sigma})\in\mathbb{R}^{n\times n}}\sum_{i=1}^n|P_i - \tilde{P}_i|^2.$$

Does anyone have references where I can learn more about this problem; for instance, whether the objective function is convex?

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  • $\begingroup$ I believe you can solve this exactly (that is, the minimum is $0$) by assuming $\sigma_i^2=1$ for all $i.$ An induction on $n$ ought to work, so--to understand the issue--start with the case $n=2.$ Indeed, you don't need many details: a topological argument will do, relying simply on the continuity of the function that maps the means $(\mu_i)$ to the probabilities $(\tilde P_i).$ $\endgroup$ – whuber Nov 28 '20 at 16:23
  • $\begingroup$ I was thinking actually that it's not well-posed even with fixed unit variances. My reasoning is the following: for any solution $\mu^\star$ if we translate by a vector $(a,\ldots,a)$ then the probabilities $\tilde{P}_i$ would all remain the same. In fact, it seems to me that $\tilde{P}_i$ is translation invariant in $\vec{\mu}$ even without unit variance. $\endgroup$ – Dennis Nov 28 '20 at 16:52
  • $\begingroup$ It is translation invariant as well as scale invariant -- but that doesn't make it ill-posed; it only means the solution is not unique. Indeed, the problem doesn't require normality: you may replace the Normal family with any location-scale family based on a distribution having an everywhere continuous density function. $\endgroup$ – whuber Nov 28 '20 at 17:30
  • $\begingroup$ For the $n = 2$ case we can evaluate $\tilde{P}_i$, but this seems non-trivial for the case where $n > 2$, is that right? If you could upvote the question it would be appreciated. $\endgroup$ – Dennis Nov 28 '20 at 19:20
  • $\begingroup$ The point is that you don't need to evaluate anything: the result is very general. $\endgroup$ – whuber Nov 28 '20 at 19:29

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